A uniform horizontal rod of mass M and length l rotates with angular velocity w (omega) about a vertical axis through its center. Attached to each end of the rod is a small mass m. Determine the angular momentum of the system about the axis. Write you're answer in terms of M, m, l and w (omega).
The moment of inertia of the rod with the two masses at each end is:
I = M*L^2/12 + 2*m*(L/2)^2
= M*L^2/12 + m*L^2/2
= (M + 6m)*L^2/12
The angular momentum is
I*w = (M+6m)*w*L^2/12.
The angular momentum of the system about the axis can be calculated by considering the individual contributions of the rod and the masses.
1. Angular momentum of the rod:
The rod can be treated as a point mass located at its center of mass. The center of mass of the rod is at a distance of l/2 from the axis of rotation. The angular momentum of the rod is given by:
L_rod = (M * (l/2) ) * w
2. Angular momentum of the masses:
Each mass m contributes to the angular momentum of the system. Since the masses are attached to the ends of the rod, their distances from the axis are equal to l/2.
The angular momentum of each mass is given by:
L_mass = (m * (l/2) ) * w
Since there are two masses, the total angular momentum contributed by the masses is:
L_masses = 2 * L_mass = 2 * (m * (l/2) ) * w = m * l * w
3. Total angular momentum of the system:
The total angular momentum of the system is the sum of the angular momenta of the rod and the masses:
L_total = L_rod + L_masses
= (M * (l/2) ) * w + m * l * w
Simplifying this expression, we can factor out w:
L_total = w * [M * (l/2) + m * l]
Therefore, the angular momentum of the system about the axis is w * [M * (l/2) + m * l].
To determine the angular momentum of the system about the axis, we can consider the angular momentum of each component separately and then sum them up.
The angular momentum of the rod can be calculated as the product of its moment of inertia and angular velocity. The moment of inertia of a uniform rod rotating about its center is given by:
I_rod = (1/12) * M * l^2
The angular momentum of the rod is then:
L_rod = I_rod * w
Now, let's consider the masses m attached to each end of the rod. The distance from the center of the rod to each mass is l/2. The moment of inertia of each mass about the axis is given by:
I_mass = m * (l/2)^2 = m * (l^2/4)
Since there are two masses, the total moment of inertia for the masses is:
I_total_masses = 2 * I_mass = m * (l^2/2)
The angular momentum of the masses is then:
L_masses = I_total_masses * w
Finally, the total angular momentum of the system about the axis is the sum of the angular momentum of the rod and the angular momentum of the masses:
L_total = L_rod + L_masses
Substituting the expressions for the moment of inertia of the rod and the masses:
L_total = (1/12) * M * l^2 * w + m * (l^2/2) * w
Therefore, the angular momentum of the system about the axis is (1/12) * M * l^2 * w + m * (l^2/2) * w, written in terms of M, m, l, and w.